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2014-06-07T14:29:48+03:00
Din teorema sinusurilor avem:

\dfrac{a}{sinA}=2R\Rightarrow sinA=\dfrac{a}{2R}\ si \ analoagele;

Din teorema cosinusului:

cos\ a=\dfrac{b^2+c^2-a^2}{2bc}; \ \ si \ analoagele

ctgA=\dfrac{cosA}{sinA}=\dfrac{b^2+c^2-a^2}{2bc}\cdot\dfrac{2R}{a}=\dfrac{R}{abc}(b^2+c^2-a^2);\ si \ analoagele

Inlocuim in relatia data si avem:

\dfrac{a^2}{ctgB+ctgC}=\dfrac{a^2}{\dfrac{R}{abc}(a^2+c^2-b^2)+\dfrac{R}{abc}(a^2+b^2-c^2)}=

=\dfrac{a^2\cdot abc}{R(a^2+c^2-b^2+a^2+b^2-c^2)}=\dfrac{abc}{2R}=2\cdot\dfrac{abc}{4R}=2S
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